Báo cáo toán học: "Chromatic Graphs, Ramsey Numbers and the Flexible Atom Conjecture" potx

Maddux, Jacob Manske Department of Mathematics University of Dallas, Irving, TX, USA alm@udallas.edu Department of Mathematics Iowa State University, Ames, IA, USA maddux@iastate.edu Dep

Chromatic Graphs, Ramsey Numbers and the Flexible

Atom Conjecture

Jeremy F Alm, Roger D Maddux, Jacob Manske

Department of Mathematics University of Dallas, Irving, TX, USA

alm@udallas.edu Department of Mathematics Iowa State University, Ames, IA, USA

maddux@iastate.edu Department of Mathematics Iowa State University, Ames, IA, USA

jmanske@iastate.edu Submitted: Nov 13, 2007; Accepted: Mar 17, 2008; Published: Mar 27, 2008

Mathematics Subject Classification: 05D40, 03G15

Abstract Let KN denote the complete graph on N vertices with vertex set V = V (KN) and edge set E = E(KN) For x, y ∈ V , let xy denote the edge between the two vertices x and y Let L be any finite set and M ⊆ L3 Let c : E → L Let [n] denote the integer set {1, 2, , n}

For x, y, z ∈ V , let c(xyz) denote the ordered triple (c(xy), c(yz), c(xz)) We say that c is good with respect to M if the following conditions obtain:

(i) ∀x, y ∈ V and ∀(c(xy), j, k) ∈ M, ∃z ∈ V such that c(xyz) = (c(xy), j, k);

(ii) ∀x, y, z ∈ V , c(xyz) ∈ M; and

(iii) ∀x ∈ V ∀` ∈ L ∃ y ∈ V such that c(xy) = `

We investigate particular subsets M ⊆ L3 and those edge colorings of KN which are good with respect to these subsets M We also remark on the connections of these subsets and colorings to projective planes, Ramsey theory, and representations

of relation algebras In particular, we prove a special case of the flexible atom conjecture

1 Motivation and background

Let KN denote the complete graph on N vertices with vertex set V = V (KN) and edge set E = E(KN) For x, y ∈ V , let xy denote the edge between the two vertices x and

y Let L be any finite set and M ⊆ L3 Let c : E → L Let [n] denote the integer set {1, 2, , n}

For x, y, z ∈ V , let c(xyz) denote the ordered triple (c(xy), c(yz), c(xz)) We say that

c is good with respect to M if the following conditions obtain:

(i) ∀x, y ∈ V and ∀(c(xy), j, k) ∈ M, ∃z ∈ V such that c(xyz) = (c(xy), j, k);

(ii) ∀x, y, z ∈ V , c(xyz) ∈ M; and

(iii) ∀x ∈ V ∀` ∈ L ∃ y ∈ V such that c(xy) = `

If K = KN has a coloring c which is good with respect to M, then we say that K realizes M (or that M is realizable)

If we take Rα = {(x, y) : c(xy) = α}, and let | stand for ordinary composition of binary relations, ie Rα|Rβ := {(x, z) : ∃y (x, y) ∈ Rα, (y, z) ∈ Rβ}, then conditions (i) and (ii) imply

(Rα|Rβ) ∩ Rγ 6= ∅ =⇒ Rγ ⊆ Rα|Rβ Conditions (i) - (iii) are given in [1] where the author calls a coloring on KN that realizes some M a symmetric color scheme It is proved in [2] that if M is a set of triples that is closed under permutation such that there is at least one α ∈ L such that for all β, γ ∈ L, (α, β, γ) ∈ M, then M is realized by a coloring on Kω, the complete graph on countably many vertices Any such color α is called a flexible color, since it can participate in any triple

Conditions (i) - (iii) may seem quite stringent, but in fact these conditions are sat-isfied in many natural situations Recall the notation for the Ramsey numbers; that is, R(k1, k2, , k`) is the minimum integer n such that in any `-coloring of the edges of

Kn there is a monochromatic complete graph on kj vertices in color j for some j In particular, the coloring of K5 which shows R(3, 3) ≥ 6 satisfies (i) - (iii), as does the coloring of K8 which shows R(4, 3) ≥ 9, the colorings of K16 that show R(3, 3, 3) ≥ 17, both “twisted” and “untwisted”, and the coloring of K29 given in [7] and [4] that shows that R(4, 3, 3) ≥ 30 In fact, the coloring of K5 without monochromatic triangles is a re-alization of M0 = {(r, b, b), (b, r, b), (b, b, r), (r, r, b), (r, b, r), (b, r, r)} The coloring of K8

mentioned above is a realization of M = M0∪ {(r, r, r)}; the col orings of K16 are real-izations of M = {r, b, g}3\{(r, r, r), (b, b, b), (g, g, g)}; the coloring of K29 is a realization

of M = {r, b, g}3\{(b, b, b), (g, g, g)}

In [1], Comer introduces the number r(k) which is the largest N such that there is a coloring on KN that realizes

M =r1, , rk}3\{(ri, ri, ri) : i ∈ [k]}

Clearly, r(k) ≤ R(

k times

z }| {

3, 3, , 3) − 1; equality holds for k = 2 and k = 3 An interesting open problem is whether equality holds for all values of k

Realizations of color schemes arise in connection with projective planes as well Let

L = {r1, , r`}, and let

M` = {(ri, rj, rk) : |{i, j, k}| ∈ {1, 3}} Lyndon proved in [5] that M` is realizable in some complete graph if and only if there exists a projective plane of order `−1, for ` > 2 This result has been extremely important

in the theory of relation algebras

In [3], Maddux, Jipsen and Tuza show that for M = L3, KN realizes M for arbitrarily large finite N In the case when M = L3, every color in L is a flexible color

2 The Main Result

The principal result of this paper is that Mn is realizable in KN for some N < ω, where

L = {r, b1, , bn} and

Mn:= {(r, r, r), (r, r, bi), (r, bi, r), (bi, r, r), (r, bi, bj), (bi, r, bj), (bi, bj, r) : i, j ∈ [n]} (Observe that Mn = {r, b1, , bn}3\ {b1, , bn}3.) This is a special case of a problem that has come to be known as the flexible atom conjecture This problem originates in relation algebra; an explanation of the conjecture in this context can be found in [6]

Theorem 1 ∀n ≥ 1 ∃r = r(n) such that ∀k > r, KN realizes Mn for N =3k − 4

k



The proof will proceed as follows First we will construct realizations of M1 in KN

for arbitrarily large N These colorings of KN will exhibit quite a lot of redundancy; in particular, for any given edge xy ∈ E and triple (c(xy), j, k) ∈ M1, there exist many vertices z such that c(xyz) = (c(xy), j, k), while condition (i) only requires that there

be one such vertex The graph KN, which is colored in colors r and b, can then be recolored by assigning edges colored b to a color from {b1, , bn} uniformly at random The probability that this recoloring realizes Mn is shown to be nonzero for sufficiently large N

Note that r is a flexible color in Mn In the case that a flexible color is present, it

is not hard to see that condition (iii) is automatically satisfied whenever (i) and (ii) are, and so we make no further mention of it

3 Proof of theorem 1

Let k ∈ N and let [3k − 4]k denote the collection of k-subsets of [3k − 4] Let G be the complete graph with vertex set V = [3k − 4]k

Lemma 1 G realizes M1 for any k ≥ 3

Proof of lemma 1 Define an edge coloring c : E(G) → {r, b} by

c(xy) =

(

b, if 0 ≤ |x ∩ y| ≤ 1,

r, otherwise

Let Er = {xy ∈ E(G) : c(xy) = r} and Eb = {xy ∈ E(G) : c(xy) = b} The following five claims establish that c satisfies condition (i) for M1

Let xy ∈ Er Since |x ∩ y| ≥ 2, |x ∪ y| ≤ 2k − 2

Claim 1 ∃z ∈ V such that c(xyz) = (r, r, r)

Let (x ∪ y) denote [3k − 4] \ (x ∪ y) and let ` be any subset of (x ∪ y) with k − |x ∩ y| elements Set z = ` ∪ (x ∩ y) We have |x ∩ z| ≥ 2 and |y ∩ z| ≥ 2, so c(xyz) = (r, r, r) and claim 1 is true

Claim 2 ∃z ∈ V such that c(xyz) = (r, r, b)

Let a1 ∈ y \ x, a2 ∈ x ∩ y, and ` be any (k − 2)-subset of (x ∪ y) Set z = ` ∪ {a1, a2}

We have |x ∩ z| = 1 and |y ∩ z| = 2, so c(xyz) = (r, r, b) and claim 2 is true

Claim 3 ∃z ∈ V such that c(xyz) = (r, b, b)

Let a1 ∈ x \ y, a2 ∈ y \ x Let ` be as in the the proof of claim 2 Set z = ` ∪ {a1, a2}

We have |x ∩ z| = |y ∩ z| = 1, so c(xyz) = (r, b, b) and claim 3 is true

Now let xy ∈ Eb Since |x ∩ y| ≤ 1, |x ∪ y| ≥ k − 3

Claim 4 ∃z ∈ V such that c(xyz) = (b, r, r)

If k = 3, then |x ∩ y| = 1, so we can pick z to be the 3-subset consisting of x ∩ y, one point from x \ y and one point in y \ x For k ≥ 4, let `1 be any 2-subset of x \ y, `2 be any 2-subset of y \ x, and `3 be any (k − 4)-subset of (x ∪ y) Set z = `1∪ `2∪ `3 We have |x ∩ z| = 2 and |y ∩ z| = 2, so c(xyz) = (b, r, r) and claim 4 is true

Claim 5 ∃z ∈ V such that c(xyz) = (b, b, r)

If k = 3, then |x ∩ y| = 1, so we can pick z to be the 3-subset consisting of y \ x together with one point from x \ y For k ≥ 4, let `1 be any 3-subset of x \ y and a ∈ y \ x Let `3 be as in the proof of claim 4 Set z = `1∪ {a} ∪ `3 We have |x ∩ z| ≥ 2 and

|y ∩ z| = 1, so c(xyz) = (b, b, r) and claim 5 is true

Observe that claims 1-5 imply that c satisfies condition (i) for M1 It remains to show that c satisfies condition (ii) for M1, which we show in claim 6 below

Claim 6 ∀x, y, z ∈ V , c(xyz) ∈ M1

By way of contradiction, suppose ∃x, y, z ∈ V with c(xyz) = (b, b, b) Since |x∪y ∪z| ≤ 3k − 4, the pigeonhole principle implies that one of |x ∩ y|, |x ∩ z|, or |y ∩ z| is greater than or equal to 2, a contradiction

Observe that claims 1-6 imply that c is good with respect to M1, and thus G realizes

M1

Let n ≥ 2 and let Er and Eb be as in the proof of lemma 1 Let S be the set of all n-ary sequences of length m = |Eb| taking digits from [n] Choose a sequence s from S

at random Enumerate the edges of Eb : e1, , em Let s(j) ∈ [n] denote the jth position

of the sequence s Define a partition of Eb into n (possibly empty) parts Eb 1, , Eb n as follows:

Eb i = {ej : s(j) = i}, i ∈ [n]

Define a new edge coloring of G given by

c0(xy) =

(

bi if xy ∈ Ebi,

r if xy ∈ Er

It is not hard to see that the probability that a given edge has color i is 1/n; and furthermore that, given two distinct edges, the assignment of their colors is independent

We claim that for sufficiently large k, c0 is good with respect to Mn, and thus G realizes Mn; for this reason, we assume that k ≥ 4 Since c satisfies condition (ii) for

M1, it is easy to see c0 satisfies condition (ii) for Mn We show that the probability that

c0 does not satisfy condition (i) for Mn is less than 1

Claim 7 The probability P1 that given xy ∈ Er, ∃i, j ∈ [n] such that ∀z ∈ V c0(xyz) 6= (r, bi, bj) is bounded from above by n2(1 − 1/n2)(k−2) 2

Proof of claim 7 Let Z := {z ∈ V : c(xyz) = (r, b, b)} For fixed i, j ∈ [n] and z ∈ Z, the probability

(xz ∈ Eb i) ∧ yz ∈ Eb j

is 1/n2, so the probability

(xz /∈ Eb i) ∨ yz /∈ Eb j

is 1 − 1/n2 Considering all z ∈ Z, we have that the probability

^

z∈Z

 (xz /∈ Ebi) ∨ yz /∈ Ebj


is (1 − 1/n2)|Z| Summing over all n2 combinations of i and j, we arrive at

P1 = n2 1 − 1/n2
|Z| (1) For an upper bound on P1 we compute a lower bound on |Z| Since we seek a lower bound, we may assume |x ∩ y| = 2 Note that |(x ∪ y)| = k − 2 Let ax ∈ x \ y and

ay ∈ y \ x If z = (x ∪ y) ∪ {ax, ay}, then z ∈ Z Since there are (k − 2)2 distinct z of this form, (k − 2)2 ≤ |Z| This fact together with (1) gives P1 ≤ n2(1 − 1/n2)(k−2)2, as desired

Claim 8 The probability P2 that given xy ∈ Er, ∃j ∈ [n] such that ∀z ∈ V c0(xyz) 6= (r, r, bj) is bounded from above by n (1 − 1/n)(k−22 )

Proof of claim 8 Let Z := {z ∈ V : c(xyz) = (r, r, b)} For fixed j ∈ [n] and z ∈ Z, the probability

(xz ∈ Eb j) ∧ (yz ∈ Er) = (xz ∈ Eb j)

is 1/n, so the probability

(xz /∈ Eb j)

is 1 − 1/n Considering all z ∈ Z, we have that the probability

^

z∈Z

(xz /∈ Eb j)

is (1 − 1/n)|Z| Summing over all j, we arrive at

P2 = n(1 − 1/n)|Z| (2) For an upper bound on P2, we compute a lower bound on |Z| As in claim 7, we may assume |x ∩ y| = 2 so |(x ∪ y)| = k − 2 Let ` be any 2-subset of (x ∪ y) If z = (y \ x) ∪ `, then z ∈ Z Since there are k − 2

2

 distinct z of this form, k − 2

2



≤ |Z| This fact together with (2) gives P2 ≤ n (1 − 1/n)(k−22 ), as desired

Claim 9 The probability P3 that given i ∈ [n] and xy ∈ Eb i, ∃j ∈ [n] such that ∀z ∈ V ,

c0(xyz) 6= (bi, r, bj) is bounded from above by n (1 − 1/n)(k4)

Proof of claim 9 Fix i ∈ [n] and xy ∈ Ebi Let

Z := {z ∈ V : c(yz) = r and c(xz) = b}

For j ∈ [n], the probability that xz ∈ Eb j is 1/n, so the probability that xz /∈ Eb j is

1 − 1/n Continuing as in claim 8, we have

P3 = n(1 − 1/n)|Z| (3) Again, we seek a lower bound for |Z|, so we may assume |x ∩ y| = 0 Note that

|(x ∪ y)| = k − 4 Let ` be any 4-subset of y If z = (x ∪ y) ∪ `, then z ∈ Z Since there are k

4

 distinct z of this form, k

4



≤ |Z| This fact together with (3) gives

P3 ≤ n(1 − 1/n)(k4)

Observe that ∀`, P1 ≥ P` Hence, we can use the upper bound in claim 7 for P1 as

an upper bound for the probability that condition (i) does not obtain for given xy ∈ E Since G has less than 3k − 4

k

2

edges, an upper bound for the probability P that c0 fails

to satisfy condition (i) for Mn is

P ≤X

e∈E

P1 ≤3k − 4

k

2

P1 ≤3k − 4

k

2

n2



1 − 1

n2

(k−2) 2

Next, we show that the right hand side of the expression in (4) can be made less than

1 by choosing k large enough Since 1 − x ≤ e−x for all x, we have

3k − 4 k

2

n2



1 − 1

n2

(k−2) 2

≤ 3k − 4

k

2

n2e−(k−2)2/n2

≤ 23k−4
2n2e−(k−2)2/n2

≤ 26kn2e−(k−2)2/n2 (5) Note that the expression in (5) is less than 1 if and only if logh26kn2

e−(k−2) 2 /n 2i

< 0, which holds just in case

6k log 2 + 2 log n − (k − 2)

2

To ensure that the inequality in (6) will hold, we first assume that k = cn2 for some

c ∈ R and realize the above as a quadratic polynomial in c Since the coefficient of c2 is negative, the function is concave down By finding the zeros of this polynomial in terms

of n and then maximizing (over n) the greatest of them, we can find the c which will guarantee the inequality in (6) For n ≥ 2, it is sufficient to take c ≥ 5.2

For such k, we have that P < 1, so there exists an edge coloring c : E(G) → {r, b1, , bn} which is good with respect to Mn Hence, G realizes Mn and the proof of theorem 1 is complete

Corollary 1 Any finite integral symmetric relation algebra with one flexible atom and with all (mandatory) diversity cycles involving the flexible atom is representable on arbi-trarily large finite sets

It is possible to obtain Corollary 1 with “symmetric” deleted in the following way We alter the proof of Theorem 1 to construct n + 1 binary relations instead of an edge-colored graph in n + 1 colors Referring to the graph colored in two colors constructed lemma 1, let R = {(x, y) : c(xy) = r} and B = {(x, y) : c(xy) = b} Partition B into n disjoint subsets in the following way: Let 2` be the number of asymmetric diversity atoms, so that b1, , b2` are asymmetric and b2`+1, , bn are all symmetric Order the vertices

v1, , vN Let V<= {(vi, vj) : i < j} Assign pairs from V< to sets B1, , Bn uniformly

at random Now we assign the remaining pairs (vj, vi) as follows For i, j with i < j, (i) if (vi, vj) ∈ Bm and m > 2`, then (vj, vi) ∈ Bm;

(ii) if (vi, vj) ∈ Bm and 1 ≤ m ≤ `, then (vj, vi) ∈ Bm+`;

(iii) if (vi, vj) ∈ Bm and ` < m ≤ 2`, then (vj, vi) ∈ Bm−`

Thus we have Bm˘ = Bm+` for m ≤ ` Then by making superficial changes to the remainder of the proof we establish the result for the nonsymmetric case Thus we have the following:

Theorem 2 Any finite integral relation algebra with one flexible atom and with all (mandatory) diversity cycles involving the flexible atom is representable on arbitrarily large finite sets

Acknowledgments

We thank Maria Axenovich for her valuable suggestions and comments, and the anony-mous reviewer for suggesting improvements to the paper We dedicate this paper to Samantha Alm, who during its writing came into existence

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